In the circuit shown above, the capacitance is C = 0.57 μF and the battery volt

Question

In the circuit shown above, the capacitance is C = 0.57 μF and the battery voltage is 9.4 V. Switch S is closed at time t=0.

a) What is the current Io leaving the battery at t=0, immediately after the switch is closed?

I =

b) What is the current I a "long time" later?

I =

c) What charge has accumulated on the capacitor after this long time?

Q =

d) If, finally, switch S is opened again, how long will it take after the switch is opened for the capacitor to lose 80% of its charge?

T =

Please include solution as well as answer.

Explanation / Answer

a)

after immediately switch is closed at t=0 ,the capacitor acts as short circuit ,so

I=E/20 =9.4/20

I=0.47 A

b)

After long time capacitor acts open circuit ,so

I=E/(20+40) =9.4/60

I=0.1567 A

c)

Q=CV =9.4*0.57*10^-6

Q=5.36*10^-6 C or 5.36 uC

d)

given

Q(t)=0.2Qo

Q(t) =Qoe^-(t/RC)

0.2Qo =Qoe^(-t/40*0.57*10^-6)

0.2=e^(-t/2.28*10^-5)

ln(0.2)=-t/2.28*10^-5

t=3.67*10^-5 s or 36.7 us

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